Confused about this problem

Question

When (1.000x10^1) grams of calcium carbonate (CaCO₃) is reacted completely with 50.0 mL of hydrochloric acid (HCl, d = 1.096 g/mL), all of calcium carbonate reacts to form a solution that weighs (5.56x10^1) g and some carbon dioxide gas (CO₂, d = 1.798 g/L) bubbles away. How many milliliters (mL) of carbon dioxide is produced? 

Answer 1

Looking at the balanced equation for this reaction, we have...

CaCO₃ + 2HCl ==> CaCl₂ + CO₂ + H₂O ... balanced equation

10 g.........50 ml..........................55.6 g........

Converting g to mol and finding the limiting reactant, we have...

10 g CaCO3 x 1 mol CaCO3 / 100g = 0.10 mols

50.0 ml HCl x 1.096 g/ml x 1 mol / 36.5 g = 1.50 mols

Since the mol ratio in the balanced eq. is 2 mol HCl : 1 mol CaCO₃ HCl is limiting

Not sure how the problem can state that all of the calcium carbonate reacts because HCl is limiting so not all of the CaCO₃ can react.

If it all reacted, by conservation of mass, you would have 10 g + 54.8 g = 64.8 g

Because the mass of the resulting solution is only 55.6 g, that would mean you lost 10 g which would be CO2

To convert this to volume, use the density...

10 g CO2 x 1 L / 1.798 g = 5.56 L